What is the Hilbert curve's equation?!
As pointed out by almagest, there is a formula for Hilbert's space filling curve in Space-Filling Curves by Hans Sagan. The following formula appears as formula 2.4.3 on page 18 of the text. If we write $t\in[0,1)$ in its base four expansion, $$t=0_{\dot 4}q_1q_2q_3\ldots,$$ then $$h(t) = \sum_{j=1}^{\infty} \frac{(-1)^{e_{0j}}}{2^j}\text{sign}(q_j) \left( \begin{array}{c} (1-d_j)q_j - 1 \\ 1-d_jq_j \end{array} \right), $$ where $e_{kj}$ denotes the number of $k$s preceding $q_j$ and $d_j=e_{0j}+e_{3j} \mod 2$.
While it takes a bit to digest, it's not too outrageously complicated and quite easy to program. If the denominator of $t$ is a power of 2, then the base four expansion will terminate and the computation is exact. For any other rational number, the base four expansion is periodic and the value can still be computed exactly using a geometric series. For irrational numbers, the sum can be truncated to obtain a good decimal approximation.
Consider, for example, your question about $t=4/7$. We have the base four expansion $$\frac{4}{7} = 0_{\dot 4}\overline{210}.$$ Now, $$ h(0_{\dot 4}210) = \left( \begin{array}{c} 1/2 \\ 1/2 \\ \end{array} \right) + \left( \begin{array}{c} 0 \\ 1/4 \\ \end{array} \right) + \left( \begin{array}{c} 0 \\ 0 \\ \end{array} \right) = \left( \begin{array}{c} 1/2 \\ 3/4 \\ \end{array} \right), $$ $$ h(0_{\dot 4}000210) = \left( \begin{array}{c} 1/16 \\ 1/16 \\ \end{array} \right) + \left( \begin{array}{c} 1/32 \\ 0 \\ \end{array} \right) + \left( \begin{array}{c} 0 \\ 0 \\ \end{array} \right) = \left( \begin{array}{c} 3/32 \\ 1/16 \\ \end{array} \right), $$ $$ h(0_{\dot 4}000000210) = \left( \begin{array}{c} 1/128 \\ 1/128 \\ \end{array} \right) + \left( \begin{array}{c} 0 \\ 1/256 \\ \end{array} \right) + \left( \begin{array}{c} 0 \\ 0 \\ \end{array} \right) = \left( \begin{array}{c} 1/128 \\ 3/256 \\ \end{array} \right), $$ and $$ h(0_{\dot 4}000000000210) = \left( \begin{array}{c} 1/1024 \\ 1/1024 \\ \end{array} \right) + \left( \begin{array}{c} 1/2048 \\ 0 \\ \end{array} \right) + \left( \begin{array}{c} 0 \\ 0 \\ \end{array} \right) = \left( \begin{array}{c} 3/2048 \\ 1/1024 \\ \end{array} \right). $$ Looking at the columns, it's not too hard to spot the pattern and we have $$ h(4/7) = \left( \begin{array}{c} 1/2 \\ 1/2 \\ \end{array} \right) \sum _{k=0}^{\infty } \frac{1}{8^k}+\left( \begin{array}{c} 1/32 \\ 1/4 \\ \end{array} \right) \sum _{k=0}^{\infty } \frac{1}{64^k} = \left( \begin{array}{c} 38/63 \\ 52/63 \\ \end{array} \right). $$
Finally, here are some approximations to Hilbert's curve obtained by passing a polygonal curve the points $h(p/2^k)$ for increasing values of $k$.
